What is the sum from n=0 to infinity of 7(-1/8)^n ?

The sum from n=0 to infinity of 7(-1/8)^n is 56/9

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sum from n=0 to infinity of 7(-1/8)^n

n = 0 \sum \infty 7 (- \frac{1}{8})^{n}

\frac{56}{9}

Solution steps

n = 0 \sum \infty 7 (- \frac{1}{8})^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 7 \cdot n = 0 \sum \infty (- \frac{1}{8})^{n}

Apply Series Geometric Test:

\frac{8}{9}

= 7 \cdot \frac{8}{9}

Simplify

7 \cdot \frac{8}{9} : \frac{56}{9}

= \frac{56}{9}

x y^{^{'}} + 4 y = 2 x^{3}

y = 7 x + 2, (2, 4)

\int e^{- 16 x} d x

\frac{d}{d x} (1 - e^{- x} sin (y))

\frac{d}{d x} (\frac{x}{x ^{2} + 5})

What is the sum from n=0 to infinity of 7(-1/8)^n ?
The sum from n=0 to infinity of 7(-1/8)^n is 56/9